an acid with the equilibrium concentration is listed below:
1) HA + H2O = H3O^+ + A-
HA= 10^-1 M, H3O=10^-3M, A=10^-3M
Calculate the Ka and pKa for the acid.
Ka=[10^-3][10^-3] / [10^-1] = 1.00 X 10^-5
pKa = -log[1.00 X 10^-5] = -1.00
Did I do this correctly for the Ka and pKa?
If that is (HA) = 0.1 before ionization, then I would have used (HA) after ionization as 0.1-0.001 = 0.099.
If Ka = 1.00E-5, then pKa = 5
pKa = -log(Ka) = -log(1E-5) = -(-5) = 5.
Yes, you have correctly calculated the Ka and pKa for the given acid.
The dissociation reaction of the acid is represented as:
HA + H2O = H3O+ + A-
From the equilibrium concentrations provided, the concentration of HA is 10^-1 M, the concentration of H3O+ is 10^-3 M, and the concentration of A- is also 10^-3 M.
To calculate the Ka (acid dissociation constant), you use the formula:
Ka = [H3O+][A-] / [HA]
Plugging in the given concentrations, we get:
Ka = (10^-3)(10^-3)/(10^-1) = 1.00 x 10^-5
So, the Ka for the acid is 1.00 x 10^-5.
To calculate the pKa (the negative logarithm of Ka), you can use the formula:
pKa = -log(Ka)
Plugging in the value of Ka, we get:
pKa = -log(1.00 x 10^-5) = -(-5) = 5
So, the pKa for the acid is 5.